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CIE 9709 2026 June Paper 33 Q7

A Level / CIE / P3

CIE 9709 2026 June Paper 33 Paper · Question 7

题目

Problem

The square roots of 3−(207)i3-(20\sqrt7)\mathrm{i} can be expressed in the Cartesian form x+iyx+\mathrm{i}y, where xx and yy are real and exact.

By first forming a quartic equation, find the square roots of 3−(207)i3-(20\sqrt7)\mathrm{i}. Give your answer in exact Cartesian form.

(5)
题目中文翻译

3−(207)i3-(20\sqrt7)\mathrm{i} 的平方根可以表示为直角坐标形式 x+iyx+\mathrm{i}y,其中 xx 和 yy 均为精确的实数。

先建立一个四次方程,再求 3−(207)i3-(20\sqrt7)\mathrm{i} 的平方根。答案写成精确的直角坐标形式。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

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平方后比较实部与虚部,消去虚部变量以得到四次方程。最后根据两个变量的乘积确定正负号搭配。

答题过程

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The equations are

x2−y2= 3,xy= −107.\begin{align*} x^2-y^2=&\,3,\\[4mm] xy=&\,-10\sqrt7. \end{align*}

As x≠0x\ne0, substituting y=−107/xy=-10\sqrt7/x gives

x2−700x2= 3,x4−3x2−700= 0,(x2−28)(x2+25)= 0.\begin{align*} x^2-\frac{700}{x^2}=&\,3,\\[4mm] x^4-3x^2-700=&\,0,\\[4mm] (x^2-28)(x^2+25)=&\,0. \end{align*}

Thus x=±27x=\pm2\sqrt7. The corresponding values of yy are ∓5\mp5. The square roots are

27−5i,−27+5i.2\sqrt7-5\mathrm{i},\qquad -2\sqrt7+5\mathrm{i}.